If two topological spaces are weak homotopy equivalent to each other, are their Cech cohomology groups the same?

## 2 Answers

$$T = \left\{ \left( x, \sin \frac{1}{x} \right ) : x \in (0,1] \right\} \cup \{(0,y)\mid y\in[-1,1]\}$$

This has trivial homotopy groups in degrees $\ge1$ but according to Wikipedia nontrivial Čech cohomology in degree 1.

To give a more enlightening answer to the question:

Cech cohomology is *not* the same as singular cohomology. However it *is* on CW-complexes. But there is CW approximation for topological spaces and singular cohomology is a weak homotopy invariant, so Cech cohomology can't be.